Gradient, Divergence and Curl
in Curvilinear Coordinates
Although cartesian orthogonal coordinates are very intuitive and easy to use, it is often found
more convenient to work with other coordinate systems. Being able to change all variables and
expression involved in a given problem, when a different coordinate system is chosen, is one of
those skills a physicist, and even more a theoretical physicist, needs to possess.
In this lecture a general method to express any variable and expression in an arbitrary curvilinear
coordinate system will be introduced and explained. We will be mainly interested to find out general expressions for the gradient, the divergence and the curl of scalar and vector fields. Specific
applications to the widely used cylindrical and spherical systems will conclude this lecture.
1
The concept of orthogonal curvilinear coordinates
The cartesian orthogonal coordinate system is very intuitive and easy to handle. Once an origin
has been fixed in space and three orthogonal scaled axis are anchored to this origin, any point
in space is uniquely determined by three real numbers, its cartesian coordinates. A curvilinear
coordinate system can be defined starting from the orthogonal cartesian one. If x, y, z are the
cartesian coordinates, the curvilinear ones, u, v, w, can be expressed as smooth functions of x,
y, z, according to:
u = u(x, y, z)
v = v(x, y, z)
(1)
w = w(x, y, z)
These functions can be inverted to give x, y, z-dependency on u, v, w:
x = x(u, v, w)
y = y(u, v, w)
z = z(u, v, w)
(2)
There are infinitely many curvilinear systems that can be defined using equations (1) and (2).
We are mostly interested in the so-called orthogonal curvilinear coordinate systems, defined as
follows. Any point in space is determined by the intersection of three “warped” planes:
u = const
,
v = const
,
w = const
We could call these surfaces as coordinate surfaces. Three curves, called coordinate curves, are
formed by the intersection of pairs of these surfaces. Accordingly, three straight lines can be
calculated as tangent lines to each coordinate curve at the space point. In an orthogonal curved
system these three tangents will be orthogonal for all points in space (see Figure 1).
In order to express differential operators, like the gradient or the divergence, in curvilinear
coordinates it is convenient to start from the infinitesimal increment in cartesian coordinates,
1
James Foadi - Oxford 2011
Figure 1: In this generic orthogonal curved coordinate system three coordinate surfaces meet at
each point P in space. Their mutual intersection gives rise to three coordinate curves which are
themselves perpendicular in P .
dr ≡ (dx, dy, dz). By considering equations (2) and expanding the differential dr, the following
equation can be obtained:
∂r
∂r
∂r
dr =
du +
dv +
dw
(3)
∂u
∂v
∂w
∂r/∂u, ∂r/∂v and ∂r/∂w are vectors tangent, respectively, to coordinate curves along u, v
and w, in P . These vectors are mutually orthogonal, because we are working with orthogonal
curvilinear coordinates. Let us call eu , ev and ew , unit-lenght vectors along ∂r/∂u, ∂r/∂v and
∂r/∂w, respectively. If we define by hu , hv and hw as:
∂r hu ≡ ∂u
,
∂r hv ≡ ∂v
,
then the infinitesimal increment (3) can be re-written as:
∂r hw ≡ ∂w dr = hu dueu + hv dvev + hw dwew
(4)
(5)
Equation (5), and associated defnitions (4), are instrumental in the derivation of many fundamental quantities used in differential calculus, when passing from a cartesian to a curvilinear
coordinate system. Let us consider, for example, polar coordinates, (r, θ), in the plane. x and y
are functions of r and θ according to:
(
x = r cos(θ)
y = r sin(θ)
To derive the correct expression for dr ≡ (dx, dy) we need first to compute hr and hθ . From (4)
we get:
∂x ∂y q
= cos2 (θ) + sin2 (θ) = 1
,
hr = ∂r ∂r Thus, dr is given by:
∂x ∂y q
= [−r sin(θ)]2 + [r cos(θ)]2 = r
,
hθ = ∂θ ∂θ dr = drer + rdθeθ
2
James Foadi - Oxford 2011
With this result we are able to derive the form of several quantities in polar coordinates. For
example, the line element is given by:
q
√
dℓ ≡ dr · dr = (dr)2 + r 2 (dθ)2
while the area element is:
dS = hr hθ drdθ = rdrdθ
For the general, 3D, case the line element is given by:
√
dℓ ≡ dr · dr = (hu du)2 + (hv dv)2 + (hw dw)2
(6)
and the volume element is:
dV ≡ [(eu · dr)eu ] · {[(ev · dr)ev ] × [(ew · dr)ew ]} = hu hv hw dudvdw
(7)
For the curl computation it is also important to have ready expressions for the surface elements
perpendicular to each coordinate curve. These elements are simply given as:
dSu = hv hw dvdw
2
,
dSv = hu hw dudw
,
dSw = hu hv dudv
(8)
Gradient in curvilinear coordinates
Given a function f (u, v, w) in a curvilinear coordinate system, we would like to find a form for
the gradient operator. In order to do so it is convenient to start from the expression for the
function differential. We have either,
df = ∇f · dr
(9)
or, expanding df using curvilinear coordinates,
df =
∂f
∂f
∂f
du +
dv +
dw
∂u
∂v
∂w
Let us now multiply and divide each term in the summation by three different functions:
df =
1 ∂f
1 ∂f
1 ∂f
hu du +
hv dv +
hw dw
hu ∂u
hv ∂v
hw ∂w
Using (5) the following identities can immediately be derived:
eu · dr = hu du
,
ev · dr = hv dv
,
ew · dr = hw dw
Thus, we can re-write df as:
1 ∂f
1 ∂f
1 ∂f
df =
eu · dr +
ev · dr +
ew · dr =
hu ∂u
hv ∂v
hw ∂w
1 ∂f
1 ∂f
1 ∂f
eu +
ev +
ew · dr
hu ∂u
hv ∂v
hw ∂w
Comparing this last result with (9) we finally get:
∇f =
1 ∂f
1 ∂f
1 ∂f
eu +
ev +
ew
hu ∂u
hv ∂v
hw ∂w
which is the general form for the gradient in curvilinear coordinates, we were looking for.
3
James Foadi - Oxford 2011
(10)
Figure 2: Volume element in curvilinear coordinates. The sides of the small parallelepiped are
given by the components of dr in equation (5). Vector v is decomposed into its u-, v- and
w-components.
3
Divergence and laplacian in curvilinear coordinates
Consider a volume element around a point P with curvilinear coordinates (u, v, w). The divergence of a vector field v at P is defined as:
1
∆V →0 ∆V
lim
I
S
v · ndS
(11)
where S is the closed surface surrounding the volume element whose volume is ∆V , and n is
the outward-pointing normal associated to the closed surface S (see Figure 2). To compute
the surface integral in equation (11) we simply have to project vector v along its u-, v- and
w-components, and multiply the result by each perpendicular area element, in turn. In Figure 2
point P is hidden in the centre of the parallelepiped. This means that P is placed at coordinates
(u, v, w), and it is half-way between u − du/2 and u + du/2 along u, half-way between v − dv/2
and v + dv/2 along v, and half-way between w − dw/2 and w + dw/2 along w. Let us compute
the contribution to the integral from vu first. This is equivalent to
−vu hv hw dvdw
where vu , hv and hw are computed at u − du/2, summed to
vu hv hw dvdw
where vu , hv and hw are computed at u + du/2. Using Taylor expansion yields the following
contribution for the vu component of the field:
Z
surf ⊥ to u
v · ndS =
∂(vu hv hw )
dudvdw
∂u
4
James Foadi - Oxford 2011
(12)
Analogous expressions are obtained considering those contribution to the integral along v and w:
Z
∂(hu vv hw )
dudvdw
∂v
(13)
surf ⊥ to v
v · ndS =
Z
∂(hu hv vw )
dudvdw
∂w
(14)
surf ⊥ to w
v · ndS =
Thus, the closed-surface integral is given by adding up (12), (13) and (14):
∂(vu hv hw ) ∂(hu vv hw ) ∂(hu hv vw )
v · ndS =
+
+
dudvdw
∂u
∂v
∂w
S
I
Using dV = hu hv hw dudvdw, we arrive, eventually, at the final expression for the divergence in
curvilinear coordinates replacing last quantity into formula (11):
1
∂(vu hv hw ) ∂(hu vv hw ) ∂(hu hv vw )
∇·v =
+
+
hu hv hw
∂u
∂v
∂w
(15)
From this last equation is also very immediate to derive the expression for the laplacian of a
scalar field ϕ. This is defined as:
∇2 ϕ ≡ ∇ · (∇ϕ)
Thus we only need to link formulas (10) and (15) together
∂
1
∇ ϕ=
hu hv hw ∂u
2
∂
∂
1 ∂ϕ
1 ∂ϕ
1 ∂ϕ
hv hw +
hw +
hu
hu hv
hu ∂u
∂v
hv ∂v
∂w
hw ∂w
to yield, eventually,
∂
1
∇ ϕ=
hu hv hw ∂u
2
4
hv hw ∂ϕ
hu ∂u
∂
+
∂v
hu hw ∂ϕ
hv ∂v
∂
+
∂w
hu hv ∂ϕ
hw ∂w
(16)
Curl in curvilinear coordinates
The curl of a vector field is another vector field. Its component along an arbitrary vector n is
given by the following expression:
1
∆S→0 ∆S
[∇ × v]n ≡ lim
I
γ
v · dr
(17)
where γ is a curve encircling the small area element ∆S, and n is perpendicular to ∆S. Let us
start with the w-component. We need to select a surface element perpendicular to ew . This is
given in Figure 3. The contribution to the line integral coming from segments 1 and 3 are
vu hu du
computed at v − dv/2, and
−vu hu du
computed at v + dv/2. These, added together, gives:
−
∂(hu vu )
dudv
∂v
The contribution from segments 2 and 4 gives, on the other hand,
vv hv dv
5
James Foadi - Oxford 2011
(18)
Figure 3: Surface element for the determination of curl’s component along w, in curvilinear
coordinates.
computed at u + du/2, and
−vv hv dv
computed at u − du/2. Adding them together yields
∂(hv vv )
dudv
∂u
(19)
From the partial results (18) and (19) we obtain the contribution to the curl we were looking for:
[∇ × v]ew =
1
∂(hv vv ) ∂(hu vu )
∂(hv vv ) ∂(hu vu )
1
−
−
dudv =
hu hv dudv
∂u
∂v
hu hv
∂u
∂v
The other two components can be derived from the previous expression with the cyclic permutation u → v → w → u. To extract all three components the following compressed determinantal
form can be used:
e
ev
ew u
1
∇×v =
(20)
∂/∂u ∂/∂v ∂/∂w hu hv hw hu vu hv vv hw vw In the final section we will derive expressions for the laplacian in cylindrical and spherical coordinates, given the importance played by Laplace’s equation in Mathematical Physics.
5
Laplacian in cylindrical and spherical coordinates
a) CYLINDRICAL COORDINATES
Cylindrical coordinates, (r, θ, z), are related
tion:


 x
y

 z
to cartesian ones through the following transforma= r cos(θ)
= r sin(θ)
=
z
(21)
Using definition (4) we get:
hr =
q
(∂x/∂r)2 + (∂y/∂r)2 + (∂z/∂r)2 =
q
cos2 (θ) + sin2 (θ) = 1
6
James Foadi - Oxford 2011
hθ =
q
q
r 2 cos2 (θ) + r 2 sin2 (θ) = r
q
√
hz = (∂x/∂z)2 + (∂y/∂z)2 + (∂z/∂z)2 = 1 = 1
(∂x/∂θ)2 + (∂y/∂θ)2 + (∂z/∂θ)2 =
Replacing these values in expression (16), yields:
1 ∂
∂ϕ
∇ ϕ=
r
r ∂r
∂r
2
∂
+
∂θ
1 ∂ϕ
r ∂θ
+
1 ∂2ϕ ∂2ϕ
+ 2
r 2 ∂θ 2
∂z
or, simplifying:
∂ϕ
1 ∂
r
∇ ϕ=
r ∂r
∂r
2
∂
∂ϕ
+
r
∂z
∂z
(22)
b) SPHERICAL COORDINATES
Spherical coordinates, (r, θ, φ), are related to cartesian coordinates through:


 x =
r sin(θ) cos(φ)
y = r cos(θ) sin(φ)

 z =
r cos(θ)
Using definition (4) we get:
hr =
hθ =
q
hφ =
q
sin2 (θ) cos2 (φ) + sin2 (θ) cos2 (φ) + cos2 (θ) = 1
(∂x/∂r)2 + (∂y/∂r)2 + (∂z/∂r)2 =
q
(∂x/∂θ)2 + (∂y/∂θ)2 + (∂z/∂θ)2 =
q
(∂x/∂φ)2
+
(∂y/∂φ)2
+
(∂z/∂φ)2
(23)
q
r 2 cos2 (θ) cos2 (φ) + r 2 cos2 (θ) cos2 (φ) + r 2 sin2 (θ) = r
=
q
r 2 sin2 (θ) sin2 (φ) + r 2 sin2 (θ) cos2 (φ) = r sin(θ)
Replacing these values in expression (16), yields:
1
∇ ϕ= 2
r sin(θ)
2
∂
1 ∂ϕ
∂ϕ
∂
∂ϕ
∂
r 2 sin(θ)
+
sin(θ)
+
∂r
∂r
∂θ
∂θ
∂φ sin(θ) ∂φ
or, simplifying:
∇2 ϕ =
1 ∂
∂ϕ
r2
2
r ∂r
∂r
+
∂2ϕ
∂
1
∂ϕ
1
sin(θ)
+
r 2 sin(θ) ∂θ
∂θ
r 2 sin2 (θ) ∂φ2
7
James Foadi - Oxford 2011
(24)